Ferrofluids, gravity, Pascal and Arquimedes.

I've found a system that seems to work in a kind of perpetual motion, BUT please, do not take it as SPAM message, and take the time to analyze the system.

Imagine a container made of two cilinders communicated with a tube. We fill this container with ferrofluid with the same density than water (for example). Now the gravity and the pressure will be in equilibrium, so the ferrofluid is equally high in both cilinders.

Now, we put a big magnet in one of the sides, so the ferrofluid level in one side goes higher than on the other side.

Now gravity, pressure and magnetic field forces are in equilibrium for the fluid.

Now, introduce a chain of buoys (every buoy weights 1Kg and has 2 Litres of volume).

Then when in ferrofluid, buoys will float, and in air, they will go down by gravity.

For the buoys, the effects of gravity and floatation will be different on one cilinder or another (force direction).

If we have different ferrofluid levels on the cilinders, the buoys will be affected in a way to rotate the chain in a continuous way!

The ferrofluid is affected by 3 fields (gravity, pressure and magnetic forces), but the buoys are only affected by gravity and floatation... so they see an imbalance, and feels different forces on one cilinder or another.

I post a drawing of the system, and some other drawings explaining how it would work.

The hard answer is: why don't you build this? If you're right then you won't need to pay for electricity again, if you're wrong then you eventually mightn't be able to afford electricity again, either way the issue is settled to my satisfaction.

The correct answer is: because you're not putting energy in (to replace the energy inevitably lost from turbulance and friction), it will slow down and stop.

Maybe the answer you're after is: The magnetic attraction counters the gravitational pull on one side, decreasing the effective weight-density of the fluid on the left side (imagine a u-shaped tube with water on one side, oil on the other). Thus, there is a smaller buoyancy force on each ball submersed on the left side than on the one submersed on the right (one on the right displaces a greater effective-weight of ferrofluid), and since the effective-weight of the fluid on each side is in equilibrium..

You can use the magnetic field to push or pull the ferrofluid. Also, you can use the ferrofluid as a magnetic "cork" or "clot" that maintains a pressure difference using magnetic forces. The buoys can pass through the ferrofluid.

Attached Files:

The buoys get their energy from somewhere, in this case from the permanent magnet used to start the loop. Eventually, this will lose its magnetism as more and more energy is given to the buoys, which of course lose energy contnuously as heat through friction etc, etc...

Simply replace the permanent magent in your setup with a magnetic solenoid and you can see that energy would be lost by the circuit as it moved the loop.

Working with magnets is not the same than working with solenoids in energy terms, under some circunstances.

Try to replace (for instance) a magnet placed on your fridge door, and do the same thing with a solenoid. You'll waste a lot of electric energy using the solenoid, but the magnet will defy gravity (almost) forever, for free.

You'll waste a lot of electric energy using the solenoid, but the magnet will defy gravity (almost) forever, for free.

I think you've fallen for a red herring. A superconducting solenoid should retain its magnetisation, without wasting electrical energy, for just as long as a good permanent magnet - the internal resistance of a cheap solenoid is not relevent here.

The buoys get their energy from somewhere, in this case from the permanent magnet used to start the loop.

The mistake here is that the buoys will not get energy. The device won't fail to run continuously; it will fail to run, period. This is because there is no net force on the loop of buoys.

One shortcut to see this is to imagine smoothly deforming this loop-of-beads shape into a continuous donut shape. The fluid will not make this torus turn, because doing so will not alter the distribution of the fluid (neither bringing the fluid on the left slightly closer to the magnet nor the fluid on the right slightly closer to the bottom).

Alternatively you can explicity calculate the forces on each individual buoy (after which you'll understand better my earlier posts), and notice they add to give zero.